1. Linear Inequalities in Two Variables 2-5
LESSON PLAN
1. Warm Up (Slide #2)
2. Objective and PA State Standards (Slide #3)
3. Vocab (Slides #4 - 5)
4. Lesson Presentation (Slide #6 – 33)
5. Text Questions (Slide #34)
6. Worksheet 2.5B (Slide #35)
7. Lesson Quiz (Slide #36,37)
Holt Algebra 2

2. Warm Up Find the intercepts of each line. 1. 3x + 2y = 18
Write the function in slope-intercept form. Then graph.
4. 2x + 3y = –3

3. Objectives Graph linear inequalities on the coordinate plane
Solve problems using linear inequalities
PA Dept. of Educ. Math Standards and Anchors:
2.8.A2.B,F
M11.D.2.1.5

4. Vocabulary
linear inequality

5. A linear inequality relates two variables using an inequality symbol such as y > 2x – 4.
Its graph is a region (…or a wedge-shaped area) of the coordinate plane bounded by a line.

6. For example:

7. Think of the underlines in the symbols ≤ and ≥ as representing solid lines on the graph.
Helpful Hint

8. Example 1A: Graphing Linear Inequalities
Graph the inequality
The boundary has a y-int. +2 and slope of .
Draw the line dashed since it is not part of the solution.
Shade the region above the boundary line to show .

9. Example 1A Continued
Check Choose a point in the solution region, such as (3, 2) and test it in the inequality. ?
2 > 1  ?
The test point satisfies the inequality, so the solution region appears to be correct.

10. Example 1B: Graphing Linear Inequalities
Graph the inequality y ≤ –1.
Recall that y= –1 is a horizontal line.
Step 1 Draw a solid line for y=–1 because the boundary line is part of the graph.
Step 2 Shade the region below the boundary line to show where y < –1. .

11. Example 1B Continued
Check The point (0, –2) is a solution because –2 ≤ –1. Note that any point on or below y = –1 is a solution, regardless of the value of x.

12. Check It Out! Example 1a
Graph the inequality y ≥ 3x –2.
The boundary line has a y–intercept of ____ and a slope of _____. -2 +3
Draw a ______ line because it is part of the solution.
Then shade the region _______ the boundary line to show y > 3x – 2.
solid
above

13. Check It Out! Example 1a Continued
Check Is (–3, 2) a solution to the inequality?
y ≥ 3x –2
2 ≥ 3(–3) –2 ?
2 ≥ (–9) –2 ?
2 > –11  ?
The test point satisfies the inequality, so the solution region appears to be correct.

14. Graph the inequality y < –3.
Check It Out! Example 1b
Graph the inequality y < –3.
Recall that y = –3 is a ________________line.
Step 1 Draw the boundary line _________ because it is not part of the solution.
Step 2 Shade the region _________ the boundary line to show where y < –3.
horizontal
dashed .
below

15. Check It Out! Example 1b Continued
Check Is (0, –4) a solution?
Yes!

16. The point (0, 0) is the easiest point to test if it is not on the boundary line.
Helpful Hint

17. Example 2: Graphing Linear Inequalities Using Intercepts
Graph 3x + 4y ≤ 12 using intercepts.
Step 1 Find the intercepts.
y-intercept (is where x = 0)
x-intercept (is where y = 0)
3x + 4y = 12
3x + 4y = 12
3(0) + 4y = 12
3x + 4(0) = 12
4y = 12
3x = 12
y = 3
x = 4

18. Graphing 3x + 4y ≤ 12 using intercepts: Step 2 Draw the boundary line.
Example 2 Continued
Graphing 3x + 4y ≤ 12 using intercepts:
Step 2 Draw the boundary line.
The line goes through (0, 3) and (4, 0). Draw a solid line for the boundary line because it is part of the graph.
(0, 3)
Step 3 Find the correct region to shade.
Substitute (0, 0) into the inequality. Because ≤ 12 is true, shade the region that contains (0, 0).
(4, 0)

19. Check It Out! Example 2
Graph 3x – 4y > 12 using intercepts.
Step 1 Find the intercepts.
Substitute x = 0 and y = 0 into 3x – 4y = 12 to find the intercepts of the boundary line.
y-intercept
x-intercept
3x – 4y = 12
3x – 4y = 12
3(0) – 4y = 12
3x – 4(0) = 12
– 4y = 12
3x = 12
y = – 3
x = 4

20. Graph 3x – 4y > 12 using intercepts. Step 2 Draw the boundary line.
Check It Out! Example 2
Graph 3x – 4y > 12 using intercepts.
Step 2 Draw the boundary line.
The line goes through (0, –3) and (4, 0). Draw the boundary line dashed because it is not part of the solution.
(4, 0)
Step 3 Find the correct region to shade.
Substitute (0, 0) into the inequality. Because >12 is false, shade the region that does not contain (0, 0).
(0, –3)

21. Many applications of inequalities in two variables use only nonnegative values for the variables. Graph only the part of the plane that includes realistic solutions.
Don’t forget which variable represents which quantity.
Caution

22. Example 3: Problem-Solving Application
A school carnival charges $4.50 for adults and $3.00 for children. The school needs to make at least $135 to cover expenses.
A. Using x as the number of adult tickets and y as the number of child tickets, write and graph an inequality for the amount the school makes on ticket sales.
B. If 25 child tickets are sold, how many adult tickets must be sold to cover expenses?

23. An inequality that models the problem is 4.5x + 3y ≥ 135.
Let x represent the number of adult tickets and y represent the number of child tickets that must be sold. Write an inequality to represent the situation.
135 y
3.00 + x
4.50
total.
is at least
number of child tickets
times
child price
plus
number of adult tickets
Adult price •
An inequality that models the problem is 4.5x + 3y ≥ 135.

24. 4.5x + 3y ≥ 135.
Find the intercepts of the boundary line.
4.5(0) + 3y = 135
4.5x + 3(0) = 135
y = 45
x = 30
Graph the boundary line through (0, 45) and (30, 0) as a solid line.
Shade the region above the line that is in the first quadrant, as ticket sales cannot be negative.

25. If 25 child tickets are sold,
4.5x + 3(25) ≥ 135
4.5x + 75 ≥ 135
4.5x ≥ 60, so x ≥ 13.3 _
At least 14 adult tickets must be sold.

26. Check It Out! Example 3
A café gives away prizes. A large prize costs the café $125, and the small prize costs $40. The café will not spend more than $1500. How many of each prize can be awarded? How many small prizes can be awarded if 4 large prizes are given away?

27. Let x represent the number of small prizes and y represent the number of large prizes, the total not too exceed $1500. Write an inequality to represent the situation.
1500 y
125 + x 40
total.
is less than
number awarded
times
large prize
plus
Small prize ≤
An inequality that models the problem is 40x + 125y ≤ 1500

28. 40x + 125y ≤ 1500
Find the intercepts of the boundary line.
40(0) + 125y = 1500
40x + 125(0) = 1500
y = 12
x = 37.5
Graph the boundary line through (0, 12) and (37.5, 0) as a solid line.
Shade the region below the line that is in the first quadrant, as prizes awarded cannot be negative.

29. 40x + 125y ≤ 1500
If 4 large prizes are awarded,
40x + 125(4) ≤ 1500
40x ≤ 1500
40x ≤ 1000
x ≤ 25
No more than 25 small prizes can be awarded.

30. > < Graphing a linear inequality with a graphing calculator.
Press the key.
Press the left arrow key to move to the left side of
Each time you press you will see one of the graph styles shown here.
>
<

31. Check It Out! Example 4 Continued
Use the calculator graph the solution to y < 4x - 4
Key “Y=“
Key “<“ twice
Key “ENTER” three times to get to the “less than” graphic
Key “>” twice to return to the equation area
Key “4X-4”
Key “GRAPH”

32. Check It Out! Example 4 Continued
Use the calculator graph the solution to
Key “Y=“
Key “<“ twice
Key “ENTER” three times
Key “>” twice
Key “(3÷4)X-3”
Key “GRAPH”

33. Graph this inequality on a calculator: 2x + 3y < -5
To use the graphing calculator, you must first rewrite the inequality so you are “solving for ‘y’”)
2x + 3y < -5
-2x
-2x
3y < -2x – 5 /3 /3 /3
y < -(2/3)x - 5/3

34. p # EVEN

36. Lesson Quiz: Part I
1. Graph 2x –5y  10 using intercepts.
2. Solve –6y < 18x – 12 for y. Graph the solution.

37. Lesson Quiz: Part II
3. Potatoes cost a chef $18 a box, and carrots cost $12 a box. The chef wants to spend no more than $144. Use x as the number of boxes of potatoes and y as the number of boxes of carrots.
a. Write an inequality for the number of boxes the chef can buy.
b. How many boxes of potatoes can the chef order if she orders 4 boxes of carrot?